Question

You have two fluids, water and a fluid that has a density that is 3/10 that of water. If a volume of water that has the same mass as

7.40 m³

of the other fluid fills a cubic aquarium, determine the pressure inside at the bottom of the aquarium. (Use 1000 kg/m³ for the density of water.)

___________ Pa

Answer 1

As per the language of the question water is filled inside the cubic aquarium, So suppose height of cubic aquarium is 'h', then Pressure at the bottom of the cube will be:

P = P_atm + rho_w*g*h

P_atm = atmospheric pressure = 1.013*10^5 N/m^2

rho_w = density of water = 1000 kg/m^3

g = acceleration due to gravity = 9.81 m/s^2

h = height of aquarium = ?

Now also given that the other fluid has density of 3/10 th of water, So

rho_1 = (3/10)*rho_w = 3*1000/10 = 300 kg/m^3

mass of 7.40 m^3 fluid has the same mass as water, So

mass of fluid = density of fluid*Volume of fluid = (300 kg/m^3)*(7.40 m^3) = 2220 kg

Now If same mass of water is filled inside the aquarium, then

Volume of aquarium = mass of water/density of water = 2220/1000 = 2.22 m^3

Now since this is cubic aquarium, So Volume = side^3

side = (Volume)^(1/3) = (2.22)^(1/3)

side = 1.3045 m

Now Pressure at the bottom of aquarium will be:

P = P_atm + rho_w*g*h

P = 1.013*10^5 + 1000*9.81*1.3045

P = 1.14*10^5 Pa

Let me know if you've any query.